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Re M06-16 [#permalink]
I think this is a high-quality question and I agree with explanation.
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Re: M06-16 [#permalink]
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@Bunuel
IF we do not take the square roots and keep solving for \(x^2 = y^2\),

­For 1) \frac{x}{y = -1 } , so x= -y, then if we put the value of x in prompt, we still get \(x^2 = y^2\). So can be True.

For 2) x = y, then it will definitely follow the prompt \(x^2 = y^2\), so this can also be True.

Can you tell me where I am thinking wrong ?

 ­
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Re: M06-16 [#permalink]
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ashish2023 wrote:
@Bunuel
IF we do not take the square roots and keep solving for \(x^2 = y^2\),

­For 1) \frac{x}{y = -1 } , so x= -y, then if we put the value of x in prompt, we still get \(x^2 = y^2\). So can be True.

For 2) x = y, then it will definitely follow the prompt \(x^2 = y^2\), so this can also be True.

Can you tell me where I am thinking wrong ?

 ­

­
While both x/y = -1 and x = y could be true, the question asks which of the followwing must be true? and those options are not necessarily true.­
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Re: M06-16 [#permalink]
a/b=x/y and a/y=b/x. This is given

From above we have:
ay=bx and
ax=by;

If we add two equations to each other;

ay+ax=bx+by;
a(y+x)=b(x+y); divide each side by (x+y);
a=b

if ax=by and a=b; then x=y. Option 2 is correct.

Please advise why this calculation is wrong
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Re: M06-16 [#permalink]
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bauyrzhan01 wrote:
a/b=x/y and a/y=b/x. This is given

From above we have:
ay=bx and
ax=by;

If we add two equations to each other;

ay+ax=bx+by;
a(y+x)=b(x+y); divide each side by (x+y);
a=b

if ax=by and a=b; then x=y. Option 2 is correct.

Please advise why this calculation is wrong

­
You cannot divide by x + y because it could be 0 and division by 0 is not allowed. For instance, consider a = 1, b = -1, x = 1, and y = -1. You should have done the following:

a(y + x) = b(x + y)

(x + y)(a - b) = 0

x + y = 0 or a - b = 0.

x + y = 0, implies x = -y.
a - b = 0, implies a = b, which gives x = y.

Hence, either x = -y or x = y. Hence, |x| = |y|.
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Re: M06-16 [#permalink]
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